// Numbas version: exam_results_page_options {"name": "Tutorial 2 - Bending moments and shear forces", "metadata": {"description": "", "licence": "Creative Commons Attribution 4.0 International"}, "duration": 0, "percentPass": "50", "showQuestionGroupNames": false, "shuffleQuestionGroups": false, "showstudentname": true, "question_groups": [{"name": "Group", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questionNames": ["", "", "", "", "", "", ""], "variable_overrides": [[], [], [], [], [], [], []], "questions": [{"name": "Determine the loadings", "extensions": [], "custom_part_types": [], "resources": [["question-resources/tutorial2_59M2sS9.svg", "/srv/numbas/media/question-resources/tutorial2_59M2sS9.svg"]], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Vladimir Vingoradov", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1862/"}], "tags": [], "metadata": {"description": "", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Shear and moment diagrams are given for the beam structure shown in the figure below. Determine the loadings on the beam. (Legth: m, Point loads: kN, Distributed loads: kN/m)

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\n

", "advice": "

Point $A$:

\n

The jump in the shear force at $A$ corresponds to the vertical reaction at $A$, so $R_A=\\var{ra}$ kN. The moment at $A$ is zero: no applied concentrated moments at $A$.

\n

Along $AB$:

\n

The shear force is constant ($dV_{AB}/dx=0$), which means there is no distributed load applied along $AB$ ($dV_{AB}/dx=-q_{AB}=0$). The moment is a linear function, whose gradient is equal to the shear force ($dM_{AB}/dx=V_{AB}=const$).

\n

Point $B$: 

\n

There is a jump of $-\\var{P}$kN in the shear force, which indicates that there is a concentration force of $\\var{P}\\mbox{kN}\\downarrow$ applied at p. $B$. The moment distribution is continuious: no concentrated moments at $B$. \\[P_B=\\var{P}\\mbox{ kN}.\\blacktriangleleft\\]

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Along $BC$:

\n

Same argument as for $AB$.

\n

Point $C$:

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Both shear force and moment diagrams are continous at $C$, no jumps: no concentrated forces and moments at $C$.

\n

Along $CD$:

\n

The shear force is a linear function (straight line); its gradient is constant and negative, which indicated that there is a uniformly distributed load acting on $CD$, whose magnitude can be determined by calculating the gradient of the function $V(x)$ ($dV/dx=-q$):
\\[q_{CD}=-\\frac{dV_{CD}}{dx}=-\\frac{-\\var{rb}-(\\var{rm})}{\\var{a}}=\\var{q}\\mbox{ kN/m}.\\,\\downarrow\\,\\blacktriangleleft\\]

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Point load (kN)Point moment (kN.m)Distributed load (kN/m)
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Along AB:

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[[0]]
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At point B:

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[[1]][[2]]
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Along BC:

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[[3]]
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At point C:

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[[4]][[5]]
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Along CD:

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[[6]]
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false, "typeendtoleave": false}, "contributors": [{"name": "Vladimir Vingoradov", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1862/"}], "tags": [], "metadata": {"description": "", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Shear and moment diagrams are given for the beam structure shown in the figure below. Determine the loadings on the beam. (Legth: m, Point loads: N, Point moments: N.m, Distributed loads: N/m)

", "advice": "

See the previous question.

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Point $A$:

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The jump in the shear force diagram implies there is a concentrated force in the downward direction \\[P_A=\\var{-ra}\\mbox{ N}.\\,\\downarrow\\,\\blacktriangleleft\\]

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The positive jump in the moment diagram means there  is a concentrated moment in the clockwise direction \\[M_A=\\var{ma}\\mbox{ Nm}.\\curvearrowright\\,\\blacktriangleleft\\]

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Along $AB$:

\n

No applied loads.

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Point $B$:

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No applied loads.

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Along $BC$:

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The shear force is a linear function, its gradient is constant \\[q_{BC}=-\\frac{dV_{BC}}{dx}=-\\frac{0-(\\var{ra})}{\\var{a2}}=-\\var{q}\\,\\mbox{ N/m}\\uparrow\\,\\blacktriangleleft\\]

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Point $C$: 

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The shear force is continuous at $C$, no concentrated forces. The bending moment distribution has a jump of $\\var{-mc}\\mbox{ Nm}$ downwards, which indicates that a concentrated moment is applied at $C$ in the counterclockwise direction.\\[M_C=\\var{mc}\\mbox{ Nm}.\\curvearrowleft\\,\\blacktriangleleft\\]

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Along $CD$:

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Same as $BC$

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\u200b

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Point load (N)Point moment (N.m)Distributed load (N/m)
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At point A:

\n
[[0]][[1]]
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Along AB:

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[[2]]
\n

At point B:

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[[3]][[4]]
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Along BC:

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[[5]]
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At point C:

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[[6]][[7]]
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Along CD:

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[[8]]
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Express the shear force $V(x)$ and moment $M(x)$ as a function of $x$ (measured from point $O$), for example \"2x+3\" or \"3*x^2+2*x-1\" or \"5x^3-12*2*x-3*5\". The functions of $x$ you enter are plotted on the graphs. (Note: positive shear force $V$ and moment $M$ should follow the agreed sign convention). Submit your answers, when you are satisfied with the graphs.

\n

Given: $L=\\var{L}$m, $w=\\var{w}$kN/m

", "advice": "

Using the differential relation $\\frac{dV}{dx}=-w$:

\n

Along $OB$ ($0 \\le x \\le \\var{L/2}$):

\n

\\[\\begin{eqnarray*}\\frac{dV}{dx}=-\\var{w},\\quad\\Rightarrow\\quad V(x)&=&V(0)+\\int_0^x (-\\var{w})dx\\\\&=&0-\\var{w}x=-\\var{w}x.\\blacktriangleleft
\\end{eqnarray*}\\]

\n

and along $BC$ ($\\var{L/2}\\le x \\le \\var{L}$):

\n

\\[\\begin{eqnarray*}\\frac{dV}{dx}=0,\\quad\\Rightarrow\\quad V(x)&=&V(\\var{L/2})+\\int_{x=\\var{L/2}}^x 0\\, dx\\\\
&=& V(\\var{L/2})=-\\var{w}\\times \\var{L/2}=-\\var{w*L/2}.\\blacktriangleleft\\end{eqnarray*}\\]

\n

Now to get the moment as a function of $x$:

\n

Along $OB$ ($0 \\le x \\le \\var{L/2}$):

\n

\\[\\begin{eqnarray*}\\frac{dM}{dx}=V(x)=-\\var{w}x,\\quad\\Rightarrow\\quad M(x)&=&M(0)+\\int_0^x (-\\var{w}x)\\,dx\\\\
&=& 0\\mbox{(free end)}-\\var{w}\\frac{x^2}{2}=-\\var{w/2} x^2.\\blacktriangleleft\\end{eqnarray*}\\]

\n

and along $BC$ ($\\var{L/2}\\le x \\le \\var{L}$):

\n

\\[\\begin{eqnarray*}\\frac{dM}{dx}=V(x)=-\\var{w*L/2},\\quad\\Rightarrow\\quad M(x)&=&M(\\var{L/2})+\\int_{\\var{L/2}}^x (-\\var{w*L/2})\\,dx\\\\
&=& -\\var{w*L^2/8}-\\var{w*L/2}(x-\\var{L/2})=\\var{w*L^2/8}-\\var{w*L/2}x.\\blacktriangleleft\\end{eqnarray*}\\]

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\n

Shear force:

\n

Along $OB$ ($0\\le x \\le \\var{L/2}$):    $V(x)=$[[0]], kN
Along $BC$ ($\\var{L/2}\\le x \\le \\var{L}$):   $V(x)=$[[1]], kN

\n

{plot(w,shear_gaps,'V, kN',L)}

\n

Bending moment:

\n

Along $OB$ ($0\\le x \\le \\var{L/2}$):  $M(x)=$[[2]], kNm
Along $BC$ ($\\var{L/2}\\le x \\le \\var{L}$): $M(x)=$[[3]], kNm

\n

{plot(w,moment_gaps,'M, kNm',L)}

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Express the shear force $V(x)$ and moment $M(x)$ as a function of $x$ (measured from point $O$), for example \"2x+3\" or \"3*x^2+2*x-1\" or \"5x^3-12*2*x-3*5\". The functions of $x$ you enter are plotted on the graphs. (Note: positive shear force $V$ and moment $M$ should follow the agreed sign convention). Submit your answers, when you are satisfied with the graphs.

\n

Given: $L=\\var{L}$m, $w=\\var{w}$kN/m, $P=\\var{P}$kN.

", "advice": "

Using the differential relation $\\frac{dV}{dx}=-w$:

\n

Along $OB$ ($0 < x < \\var{L/2}$):

\n

\\[\\begin{eqnarray*}\\frac{dV}{dx}=-\\var{w},\\quad\\Rightarrow\\quad V_{OB}(x)&=&V(0)+\\int_0^x (-\\var{w})dx\\\\&=&0-\\var{w}x=-\\var{w}x.\\blacktriangleleft
\\end{eqnarray*}\\]

\n

and along $BC$ ($\\var{L/2} < x < \\var{L}$):

\n

\\[\\begin{eqnarray*}\\frac{dV_{BC}}{dx}=0,\\quad\\Rightarrow\\quad V_{BC}(x)&=&V(\\var{L/2})+\\int_{\\var{L/2}}^x 0\\, dx\\\\
&=& V_{OB}(\\var{L/2})-P=-\\var{w}\\times \\var{L/2}-\\var{P}=-\\var{w*L/2+P}.\\blacktriangleleft\\end{eqnarray*}\\]

\n

Notice that you need to add the concentrated force $P$, which acts at p. $B$. In the shear force diagram there will be a jump in the direction of $P$.

\n

Now to get the moment as a function of $x$:

\n

Along $OB$ ($0 < x < \\var{L/2}$):

\n

\\[\\begin{eqnarray*}\\frac{dM_{OB}}{dx}=V_{OB}(x)=-\\var{w}x,\\quad\\Rightarrow\\quad M_{OB}(x)&=&M(0)+\\int_0^x (-\\var{w}x)\\,dx\\\\
&=& 0\\mbox{(free end)}-\\var{w}\\frac{x^2}{2}=-\\var{w/2} x^2.\\blacktriangleleft\\end{eqnarray*}\\]

\n

and along $BC$ ($\\var{L/2} < x < \\var{L}$):

\n

\\[\\begin{eqnarray*}\\frac{dM_{BC}}{dx}=V_{BC}(x)=-\\var{w*L/2+P},\\quad\\Rightarrow\\quad M_{BC}(x)&=&M(\\var{L/2})+\\int_{\\var{L/2}}^x (-\\var{w*L/2+P})\\,dx\\\\
&=& -\\var{w*L^2/8}\\mbox{ (no external moments at B)}-\\var{w*L/2+P}(x-\\var{L/2})=\\var{w*L^2/8+P*L/2}-\\var{w*L/2+P}x.\\blacktriangleleft\\end{eqnarray*}\\]

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\n

Shear force:

\n

Along $OB$ ($0\\le x \\le \\var{L/2}$):    $V(x)=$[[0]], kN
Along $BC$ ($\\var{L/2}\\le x \\le \\var{L}$):   $V(x)=$[[1]], kN

\n

{plot(w,shear_gaps,'V, kN',L)}

\n

Bending moment:

\n

Along $OB$ ($0\\le x \\le \\var{L/2}$):  $M(x)=$[[2]], kNm
Along $BC$ ($\\var{L/2}\\le x \\le \\var{L}$): $M(x)=$[[3]], kNm

\n

{plot(w,moment_gaps,'M, kNm',L)}

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For the beam shown in the figure below, determine the magintude of the maximum shear force and bending moment, if:

\n

", "advice": "

(a) $L=\\var{L}$ m, $P=\\var{P1}$ kN and $w=\\var{w}$ kN/m.

\n
    \n
  1. Determine the reactions at the supports:
    \\[R_{\\text{left}}=\\frac{2}{3}P+\\frac{1}{18}wL=\\var{2*P1/3+w*L/18} \\mbox{ kN},\\uparrow\\]
    \\[R_{\\text{right}}=\\frac{1}{3}P+\\frac{5}{18}wL=\\var{P1/3+5*w*L/18} \\mbox{ kN}.\\uparrow\\]
  2. \n
  3. Draw the shear force and bending moment diagrams:



  4. \n
  5. The maximum shear force is \\[|V_{\\max}|=\\var{v_max_1}\\mbox{ kN}.\\,\\blacktriangleleft\\]
  6. \n
\n

an dthe maximum bending moment is \\[M_{\\max}=\\var{m_max_1}\\mbox{ kNm}.\\,\\blacktriangleleft\\]

\n

(b) $L=\\var{L}$ m, $P=\\var{P2}$ kN and $w=\\var{w}$ kN/m.

\n
    \n
  1. Determine the reactions at the supports:
    \\[R_A=\\frac{2}{3}P+\\frac{1}{18}wL=\\var{2*P2/3+w*L/18} \\mbox{ kN},\\uparrow\\]
    \\[R_B=\\frac{1}{3}P+\\frac{5}{18}wL=\\var{P2/3+5*w*L/18} \\mbox{ kN}.\\uparrow\\]
  2. \n
  3. The shear force and bending moment diagrams:




  4. \n
  5. The maximum shear force is \\[|V_{\\max}|=\\var{v_max_2},\\mbox{ kN}.\\,\\blacktriangleleft\\]
  6. \n
  7. Firstly, determine the location  $x^*$ of the maximum bending moment:
    \\[\\frac{dM(x^*)}{dx}=V(x^*)=0,\\quad\\Rightarrow\\,\\, \\var{vb2}-\\var{w}(x^*-\\var{2*L/3})=0,\\]
    Solve for $x^*$:
    \\[x^*=\\frac{\\var{vb2}+\\var{w}\\times\\var{2*L/3}}{\\var{w}}=\\var{x_m_max_2}\\mbox{m}.\\]
    Now, calculate the moment at $x^*$. It can be done by looking at equilibrium of the beam part to the right of $x^*$:
    \\[M(x^*)=R_{\\text{right}}(L-x^*)+\\frac{w(L-x^*)^2}{2}=\\var{m_max_2}\\mbox{ kNm}.\\blacktriangleleft\\]
    Alternatively, we can integrate the shear force from $x=\\var{2*L/3}$ to $x^*$:
    \\[M(x^*)=\\var{mc2}+\\frac{1}{2}\\var{vb2}(\\var{x_m_max_2}-\\var{2*L/3})=\\var{m_max_2}\\mbox{ kNm}.\\blacktriangleleft\\]
  8. \n
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Location of the max bending moment/zero shear force

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$L=\\var{L}$ m, $P=\\var{P1}$ kN and $w=\\var{w}$ kN/m.

\n

\n

$|V_{\\max}|=$[[0]] kN, 

\n

$|M_{\\max}|=$[[1]] kNm.

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$L=\\var{L}$ m, $P=\\var{P2}$ kN and $w=\\var{w}$ kN/m.

\n

$|V_{\\max}|=$[[0]] kN,

\n

$|M_{\\max}|=$[[1]] kNm.

\n

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A simply supported steel beam of length $L=\\var{L}$ m is loaded as shown in the figure below, when $P=\\var{P[1]}$ N and $w=\\var{w[0]}$ N/m.

\n

", "advice": "

Determine the reactions:

\n

$R_A=\\var{rA}$ N.

\n

$R_D=\\var{rB}$ N

\n

(a)

\n

Shear force distribution $V(x)$, N (you can hover over a point to see its coordinates)

\n

{correctv(x,v_left,v_right,vlimits)}

\n

(b)

\n

Bending moment distribution $M(x)$, Nm:

\n

{correctm(x ,m_left,m_right,m_middle,mlimits)}

\n

(c)

\n

Since $dM(x)/dx=V(x)$, the maximum bending moment is where $V(x_{cr})=0$.

\n

Along AB the shear force is $V(x)=R_A-w x$. Hence, the shear force vanishes at 

\n

\\[x_{cr}=\\frac{R_A}{w}=\\var{xcr} \\text{m}. \\blacktriangleleft\\]

\n

The moment is the integral of the shear force (area under the shear force distribution):

\n

\\[M(x_{cr})=\\underbrace{M_A}_{=0}+\\frac{1}{2}R_A x_{cr}=0.5\\times \\var{rA}\\times \\var{xcr}=\\var{mmax}\\text{Nm}.\\blacktriangleleft\\]

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\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
PointAB leftB rightC leftC rightD
$V$, N[[0]][[1]][[2]][[3]][[4]][[5]]
\n
\n
{dragpoint_board(intervals,vlimits)}
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Fill in the table of values for $M(x)$ at the end and middle points of each interval. You can also hold and drag the red points on the graph.  \u200b

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
PointAmidle ABB leftB rightmidle BCC leftC rightmidle CDD
$M$, Nm[[0]][[1]][[2]][[3]][[4]][[5]][[6]][[7]][[8]]
\n
\n
{piecewise_moment(intervalsM,mlimits)}
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Determine magnitude of the maximum bending moment (in Nm) and its location $x_{cr}$ measured in m from point A.

\n

$M_\\max=$[[0]] 

\n

$x_{cr}=$[[1]]

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A simply supported steel beam of length $L=\\var{L}$ m is loaded as shown in the figure below, when $P_1=\\var{P[0]}$ N, $P_2=\\var{P[1]}$ N and $w=\\var{w[1]}$ N/m.

\n

", "advice": "

Determine the reactions:

\n

$R_A=\\var{rA}$ N.

\n

$R_D=\\var{rB}$ N

\n

(a)

\n

Shear force distribution $V(x)$, N (you can hover over a point to see its coordinates)

\n

{correctv(x,v_left,v_right,vlimits)}

\n

(b)

\n

Bending moment distribution $M(x)$, Nm:

\n

{correctm(x ,m_left,m_right,m_middle,mlimits)}

\n

(c)

\n

Since $dM(x)/dx=V(x)$, the maximum bending moment is where $V(x_{cr})=0$.

\n

Along BC the shear force is $V(x)=V_{B}-w x$, when $x$ is measured form B. Hence, the shear force vanishes at 

\n

\\[x_{cr}=\\frac{V_B}{w}=\\var{xcr} \\text{m},\\]

\n

Distance from A: \\[d=\\var{xcr}+\\var{L/3}=\\var{xcr+L/3}\\text{m}.\\blacktriangleleft\\]

\n

The moment is the integral of the shear force (area under the shear force distribution):

\n

\\[M(x_{cr})=M_B+\\frac{1}{2}V_B x_{cr}=\\var{m_right[1]}+0.5\\times \\var{v_right[1]}\\times \\var{xcr}=\\var{mmax}\\text{Nm}.\\blacktriangleleft\\]

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\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
PointAB leftB rightC leftC rightD
$V$, N[[0]][[1]][[2]][[3]][[4]][[5]]
\n
\n
{dragpoint_board(intervals,vlimits)}
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Fill in the table of values for $M(x)$ at the end and middle points of each interval. You can also hold and drag the red points on the graph.  \u200b

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
PointAmidle ABB leftB rightmidle BCC leftC rightmidle CDD
$M$, Nm[[0]][[1]][[2]][[3]][[4]][[5]][[6]][[7]][[8]]
\n
\n
{piecewise_moment(intervalsM,mlimits)}
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Determine magnitude of the maximum bending moment (in Nm) and its location $d$ measured from point A (in m).

\n

$M_\\max=$[[0]] 

\n

$d=$[[1]]

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